Analysis and Design of Telecommunications Systems

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Analysis and Design of Telecommunications

Systems

Marc Van Droogenbroeck

Montefiore Institute, University of Li`ege, Belgium

Academic year: 2020-2021

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Practical information

I Instructor: Marc Van Droogenbroeck

I Assistant: Renaud Vandeghen

I Evaluation

written(!): theory + exercise

I Slides:

http://orbi.uliege.be

https://orbi.uliege.be/handle/2268/170889

I New from September 2019: manual of exercises: http://hdl.handle.net/2268/239453

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General overview

In this course:

1 Theory on signals and noise

2 Study of specific aspects of telecommunications systems 3 Towards engineering rules

But why?

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Components of a telecommunication system

Main components:

1 signals

2 electronics (transmitter, receiver, connectors, repeaters, etc). 3 channel: cable (+ adapters), wireless

4 propagation issues: antennas, fading

Main constraints:

1 shared channel bandwidth

2 perturbations due to noise.

Goals of a good design:

increase the Signal to Noise ratio _NS to reach the maximal channel capacity (typically for a white Gaussian noise)

enable a communication with many users (seen as interference for the main link)

3 power consumption

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Main concerns related to signals

I Signal source handling (preparation of the signal, at the source, in the transmitter):

filtering (remove what is useless for communications) analog _{↔ digital (}digitization)

remove the redundancy in the signal: compression

I Signal over the channel:

signal shaping to make it suitable for transmission (coding, modulation, multiplexing, etc)

signal power versus the noise signal (protect the signal against noise effects)

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Outline

1 Reminder

2 Representation of bandpass signals

3 Noise in telecommunications systems

4 Digital modulation

5 Spread spectrum

6 Channels for digital communications and intersymbol

interference

7 Navigation systems

8 Multiplexing

9 Telephone traffic engineering

10 Transmission over twisted pairs (fixed telephone network)

11 Radio engineering

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Outline

1 Reminder

5 Spread spectrum

interference

8 Multiplexing

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Reminder

I Physical layer (not all the transmission protocols)

I Deterministic - stochastic processes

I A tool for characterizing stochastic processes (including noise) and linear systems: the power spectrum or power spectral density

I Properties of the power spectrum

I Gaussian process I Noise I Modulation I Digital communications 9 / 495

Network

_{⇒ protocols I}

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Network

_{⇒ protocols II}

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Deterministic vs stochastic tools

transmitter receiver User’s signal deterministic stochastic

Noise and interference stochastic stochastic

deterministic stochastic signal to consider voltage / current power

power analysis instantaneous power Power Spectral Density (PSD) p(t) = |v(t)|_R 2 = R _|i(t)|2 EX2(t) =R+∞

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Deterministic vs stochastic tools

transmitter receiver User’s signal deterministic stochastic

Noise and interference stochastic stochastic

deterministic stochastic

signal to consider voltage / current power

power analysis instantaneous power Power Spectral Density (PSD) p(t) = |v(t)|_R 2 = R _|i(t)|2 EX2(t) =R_−∞+∞γX(f )df

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More about the

P

ower

S

pectral

D

ensity (

PSD

) I

Let X (t) be a (wide-sense) stationary stochastic process (note that we use a “capital” letter X for stochastic processes).

Definition (Auto-correlation function)

ΓXX (τ ) = E {X (t + τ)X (t)} ∀t (1)

Extremely useful because it expresses the average power (when

τ = 0):

ΓXX (τ = 0) = E n

X2(t)o (2)

In practice, we have that the power PX of a stochastic process is

given by: PX = E n X2(t)o (3)

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More about the

P

ower

S

pectral

D

ensity (

PSD

) II

Definition (Power spectrum or power spectral density of a stationary process PSD) γ_X(f ) = Z ₊_∞ −∞ ΓXX (τ ) e −2πjf τ_dτ ₍₄₎ Summary: PX = E n X2(t)o (5) = E _{{X (t}+0)X (t)_{} =} Z ₊_∞ −∞ γX(f )e 2πjf0_df ₍₆₎ = Z ₊_∞ −∞ γX(f )df (7)

Therefore, γX(f ) expresses the distribution of power for all the

frequencies.

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Example of a PSD

Let us consider a signal with a random phase θ, uniformly distributed over [_{−π, +π] (or, likewise, [0, 2π])} X (t) = Ac cos (2πfct +Θ) (8) 1 _{Mean of X (t)?} µX(t) = E {X (t)} = Z +π −π Accos (2πfct +θ) 1 2π dθ = 0 (9) 2 _{Auto-correlation?} ΓXX (t1, t2) = E {X (t1)X (t2)} (10) = Z +π −π Accos (2πfct1 +θ) Accos (2πfct2 +θ) 1 2π dθ = A 2 c 2 cos [2πfc(t2 − t1)] = A2c 2 cos [2πfcτ ] (11) The signal is stationary. So,

I Power spectral density?

γX(f )= A

2 c

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Practical link between x(t) and X (t)? I

Let:

I X (t) be the stochastic (unknown) process behind the scenes,

I x(t) be the observation. This is what you measure/observe. You can estimate the PSD of X (t) by taking the Fourier transform of the observation x(t).

Proof.

Assume x(t) is deterministic and has finite energy, that is

Z +∞ −∞

|x(t)|2dt (13)

Let us define:

I a “pseudo” auto-correlation function by Γxx(τ ) =

Z +∞ −∞

x(t)x(t + τ )dt (14)

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Practical link between x(t) and X (t)? II

I and a “pseudo” PSD, which is an estimate of the true PSD, as

γx(f ) =

Z +∞ −∞

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Practical link between x(t) and X (t)? III

Some calculations lead to:

γx(f ) = Z +∞ −∞ Γxx (τ )e−2πjf τdτ (16) = Z +∞ −∞ Z +∞ −∞ x(t)x(t + τ )dt e−2πjf τdτ (17) = Z +∞ −∞ x(t) Z +∞ −∞ x(t + τ )e−2πjf τdτ dt (18) = Z +∞ −∞ x(t) _{X (f )e}2πjftdt (19) = _{X (f )} Z +∞ −∞ x(t)e2πjftdt (20) = _{X (f )}_X∗(f ) (21) = _{kX (f )k}2 (22) 19 / 495

Practical link between x(t) and X (t)? IV

In decibels (remember that v _{↔ 10 log}₁₀(v ) [dB]),

γ_x(f ) [dB] = 20 log₁₀_{kX (f )k} (23) So, γX(f ) [dB] can be estimated with the help of kX (f )k.

However,

I γ_X(f ) is not based on a single observation and therefore

γ_X(f ) is not equal to γx(f ). x [W ] 10 log₁₀(x) [dBW ] 1 [W ] 0 [dBW ] 2 [W ] 3 [dBW ] 0, 5 [W ] _{−3 [dBW ]} 5 [W ] 7 [dBW ] 10n[W ] 10_{× n [dBW ]}

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What if a process is not stationary?

Stationary (in the wide sense) means

1 the mean of X (t) is time independent:

µX(t) = µX ∀t (24)

2 the autocorrelation only depends on time differences (and not the origin of time):

ΓXX (t1, t2) = ΓXX (t2 − t1) ∀t1, t2 (25)

If a process X (t) is not stationary, then

I you cannot define a PSD for a non-stationary process.

I But, you can sometimes make it stationary. 2 classical ways worth trying:

1 _{inject a random (independent) phase, suited} _{for modulated signals}_:

S(t) = X (t) cos(2πfct +Φ) with pdf_Φ(φ) = _2π1 for φ∈ [0, 2π].

2 _{inject a random (independent) time shift (}≡jitter), suited _{for digital}

signals: X (t +T0) with pdf_T₀(t0) = _T1 for t0 ∈ [0, T ].

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Power spectral density and linear systems (= filtering)

Consider a stationary process X (t), a linear system whose transfer function is given by _H(f ), and Y (t) the output process.

Theorem (Mean of a filtered stochastic process)

Mean of Y (t):

µ_Y = µXH(0) (26)

Theorem (Wiener-Kintchine)

Power spectrum of a filtered stochastic process Y (t):

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Useful considerations about the PSDs

I Sum of (stationary) stochastic processes:

Y (t) = K (t) + N(t) (28) If both signals are uncorrelated (which they are if they are independent), then

γ_YY(f ) = γKK(f ) + γNN(f ) (29)

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Gaussian processes

What does it mean that a process X (t) is “Gaussian”?

I the probability density function (pdf) of its voltage/current is Gaussian distributed: pdf_X = fX(x) = 1 σ_X√2πe −(x−µX) 2 2σ2 X (30)

I the mean and variance of X suffice to characterize it.

I it is a good approximation for the sum of a number of

independent random variables with arbitrary one-dimensional pdfs.

Useful properties:

I If the input of a linear system is a Gaussian stochastic process, then the output is also a Gaussian process.

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Noise - white noise I

Definition (White noise)

A white noise is defined as a stochastic process whose power spectral density is constant for each frequency

γ_N(f ) = N0 2 W Hz (31) In practice, there is no “pure” white noise, but it does not matter as long as the PSD is constant inside the useful bandwidth.

A common signal in telecommunications is a wide-sense stationary zero-mean white Gaussian noise:

I the probability density function of the voltage of the noise is a Gaussian.

I the observed mean voltage has a zero mean.

I its power spectrum is constant for each frequency.

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Noise - white noise II

Fact

Power of a white noise (for a B large bandwidth) PN = N = Z ₊_∞ −∞ γN(f ) df = 2 Z _f_c₊B 2 fc−B₂ N0 2 df = 2× B × N0 2 = B N0 (32)

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Modulation

Principle: modulation is all about using of a carrier fc for

transmitting information

Amplitude

s(t) = A(t) cos (2πf (t)t + φ(t))

Phase

Frequency modulation [FM]

Phase modulation [PM]

Amplitude modulation [AM]

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Consequences of modulation

I On analog signals:

frequency band is shifted towards the carrier frequency (_{⇒ f}c)

bandwidth modification

I On digital signals:

power spectral density is shifted

γS(f ) =

1

4[γM(f − fc) + γM(f + fc)] (33) shape of the power spectral density may be modified

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Rice’s decomposition

An expression such as

s(t) = A(t) cos(2πfct + φ(t)) (34)

may also be written as

s(t) = A(t) cos(2πfct + φ(t)) (35)

= sI(t) cos(2πfct)−sQ(t) sin(2πfct)

such that

sI(t) = A(t) cos φ(t) (36)

sQ(t) = A(t) sin φ(t) (37)

Rice’s decomposition is essential because, as seen later, it means that any modulated signal can be decomposed into two amplitude modulated signals.

Many receivers in telecommunications use this principle!

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Information theory and channel capacity: there is

maximum bit rate! I

Theorem (Shannon-Hartley)

Channel capacity C (conditions for the error rate Pe → 0)

C [b/s] = B log₂ 1 + S N (38) where

I B is the channel bandwidth in Hz

I S

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On the importance of the

Eb

N0

ratio for digital transmissions

Assume infinite bandwidth and a Gaussian white channel,

C = lim B→ ∞ B log₂ 1 + S N (39) As

I S = EbRb (Eb is the energy of one bit and Rb = _T1_b is the

bitrate) I N = B N0 Therefore, C = lim B_{→ ∞} B log₂ 1 + EbRb B N0 = lim x→0    log₂1 + xEbRb N0 x    H = log2 e limx_→0 ( 1 1 +xEbRb N0 EbRb N0 ) = 1 ln 2 EbRb N0 (40)

At maximum capacity: C = Rb, so that _NEb₀ = ln 2 ≡ −1.59 [dB] is

the absolute minimum.

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Outline

1 Reminder

5 Spread spectrum

interference

8 Multiplexing

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Towards a dedicated representation for bandpass signals

Why do we use “frequency-based” representations?

[Systems] it is convenient for linear systems, and most

communication systems are linear (channel, filter, etc).

[Sharing] channels can be shared if signals are “frequency”-friendly (multiplexing).

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What type of

frequency signals

do we have?

1 in its original form, a signal is in its baseband (voice is in

[300, 3400] Hz).

2 bandwidth ≡ band occupied by the signal.

3 for digital signals, sampling −→ sampling frequency (driven by

Nyquist’s criterion and relating to the highest frequency, thus the frequency content).

4 spectrum

1 V(f ) for deterministic signals 2 γ_V(f ) for stochastic processes.

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Purposes of dealing with

bandpass

signals specifically

1 Lower the sampling frequency (not twice the highest

frequency!)

2 There are many more convenient representations of such a

signal (analytic signal, complex envelope, baseband equivalent, quadrature components, etc):

1 They are equivalent, but not equal.

2 Which one is the most appropriate depends on the context (hard to foresee).

3 See these representations as tools!

4 The many representations are convenient theoretical and practical ways to process signals.

5 Almost every receiver uses the underlying theory of bandpass signals.

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Representation of bandpass signals: outline

I Sampling of band-limited signals

I Representations of deterministic band-limited (bandpass) signals:

bandpass equivalent, analytic signal,

complex envelope, and more !

I Representation of bandpass systems

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Sampling of bandpass signals

Theorem (“Revised” sampling theorem)

Assume that v (t) is a deterministic, energy limited signal, with a W large bandwidth, whose Fourier _{V(f ) transform is} upper

bounded by fu (that is V(f ) = 0 for f > fu).

Then, it is possible to characterize this signal with samples v [nTs],

n _{∈ {−∞, +∞}, taken at a f}s sampling frequency if this frequency

is equal to fs = 2f_ku, where k is the largest integer strictly smaller

than fu

W.

It should be noted that not all sampling frequencies are valid (in order to reconstruct v (t) perfectly), except for all those that are strictly larger than 2fu.

Important practical consequences:

1 sampling frequency: from f_s = 2f_u to f_s = 2fu

k . k times less!

2 there are other ways to look at bandpass signals: new

representations!

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Representations of deterministic bandpass signals

Definition (Bandpass)

A bandpass signal v (t) is a signal if there exist two values B and f0, such that B f0, and

∀f 6∈ f0 − B 2, f0 + B 2 , _{V(f ) = 0} (41)

Definition (Equivalent baseband)

Assume a deterministic bandpass signal v (t). v (t) is an equivalent

baseband of v (t) if there exists a frequency f0, comprised inside

the frequency band of v (t), such that

v (t) = Rev (t)e2πjf0t+jϕ0 ₍₄₂₎

Note that v in fact covers a family of equivalent baseband signals (since v (t) = (v (t) + jz(t))e−2πjf0t−jϕ0 _{are all valid candidates).}

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Representations of deterministic bandpass signals

Definition (Bandpass)

A bandpass signal v (t) is a signal if there exist two values B and f0, such that B f0, and

∀f 6∈ f0 − B 2, f0 + B 2 , _{V(f ) = 0} (41)

Definition (Equivalent baseband)

Assume a deterministic bandpass signal v (t). v (t) is an equivalent baseband of v (t) if there exists a frequency f0, comprised inside

the frequency band of v (t), such that

v (t) = Rev (t)e2πjf0t+jϕ0 ₍₄₂₎

Note that v in fact covers a family of equivalent baseband signals (since v (t) = (v (t) + jz(t))e−2πjf0t−jϕ0 are all valid candidates).

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Summary of some representations of a bandpass signal

v(t) vI(t) + jvQ(t) v_a(t) e_v(t) ≡ a_v(t)ejφv(t) ≡ Re(.) ×e−2πjf0t ×e+2πjf0t ⊗ δ(t) + _πtj v_I(t) cos(.)_{− v}_Q(t) sin(.)

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Working on the spectrum I

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Working on the spectrum II

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Analytic signal I

f

2 1

||H(f)||

Let us consider the following filter that removes the negative frequency components of a signal with the (Heaviside) step function:

H(f ) =

(

0 if f < 0

2 if f _{≥ 0} = 1 + sign(f ) (43) whose impulse response is

h(t) = δ(t) + j

πt (44)

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Analytic signal II

Definition (Analytic signal)

A signal which has no negative-frequency components is called an

analytic signal. In the time domain, it is obtained as va(t) = v (t) ⊗ δ(t) + j πt (45) = v (t) + jv (t)_⊗ 1 πt (46)

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Representation map: analytic signal

v(t) v_I(t) + jv_Q(t) va(t) ev(t) ≡ av(t)ejφv(t) ≡ Re(.) ×e−2πjf0t ×e+2πjf0t ⊗ δ(t) + _πtj vI(t) cos(.)− vQ(t) sin(.) 45 / 495

Hilbert transform

Definition (Hilbert transform)

The Hilbert transform of a signal v (t), denoted as v (t), is definede

by

e

v (t) = v (t) _⊗ 1

πt (47)

With this definition,

va(t) = v (t) + jv (t)e (48)

Properties of the Hilbert transform

I The energy (or power) of a signal and that of its Hilbert transform are equal.

I [Hilbert transform of a modulated signal] Assume that v (t) is a baseband signal, then

^

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Hilbert transform

Definition (Hilbert transform)

The Hilbert transform of a signal v (t), denoted as v (t), is definede

by

e

v (t) = v (t) _⊗ 1

πt (47)

With this definition,

va(t) = v (t) + jv (t)e (48)

Properties of the Hilbert transform

I The energy (or power) of a signal and that of its Hilbert transform are equal.

I [Hilbert transform of a modulated signal] Assume that v (t) is a baseband signal, then

^ v (t) cos(2πfct) = v (t) sin(2πfct) (49) 47 / 495

Going backwards I

va(t) = v (t) + jv (t)e (50) So, v (t) = Re (va(t)) (51) v(t) vI(t) + jvQ(t) v_a(t) e_v(t) ≡ a_v(t)ejφv(t) ≡ Re(.) ×e−2πjf0t ×e+2πjf0t ⊗ δ(t) + _πtj v_I(t) cos(.)_{− v}_Q(t) sin(.)

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Going backwards II

Properties of the analytic signal I it has no negative frequency

I it carries the same power as the original signal

I v (t) can be reconstructed from va(t)

I va(t) is not a real signal. Is that a problem?

Why would we want to use va(t)?

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Baseband representation derived from the analytic signal I

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Baseband representation derived from the analytic signal II

Definition (Complex envelope of a signal)

The signal that results from a right-to-left shift of the analytic signal in the frequency domain is named the complex envelope of the signal. It is denoted as ev(t).

Mathematically, the complex envelope and its spectrum are related to the analytic signal as follows:

ev(t) = va(t)e−2πjf0t (53)

Ev(f ) = Va(f + f0) (54)

It remains to find a good practical method to determine the complex envelope!

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Representation map: complex envelope

v(t) vI(t) + jvQ(t) va(t) ev(t) ≡ av(t)ejφv(t) ≡ Re(.) ×e−2πjf0t ×e+2πjf0t ⊗ δ(t) + _πtj vI(t) cos(.)− vQ(t) sin(.) Question

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Rice components I

ev(t) = vI(t) + jvQ(t) (55) v(t) vI(t) + jvQ(t) va(t) ev(t) ≡ av(t)ejφv(t) ≡ Re(.) ×e−2πjf0t ×e+2πjf0t ⊗ δ(t) + _πtj vI(t) cos(.)− vQ(t) sin(.) 53 / 495

Theoretical determination of Rice components I

In-phase component: vI(t) vI(t) = Re (ev(t)) (56) = Reva(t)e−2πjf0t (57) = Re(v (t) + jv (t)) ee −2πjf0t ₍₅₈₎ = v (t) cos(2πf0t) + v (t) sin(2πfe 0t) (59) Quadrature component: vQ(t) vQ(t) = Im (ev(t)) (60) = Imva(t)e−2πjf0t (61) = _{−v(t) sin(2πf}0t) + v (t) cos(2πfe 0t) (62) Therefore,

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Theoretical determination of Rice components II

v (t) = Re (va(t)) (63) = Reev(t)e2πjf0t (64) = Re(vI(t) + jvQ(t))e2πjf0t (65) = vI(t) cos(2πf0t) − vQ(t) sin(2πf0t) (66) v(t) v_I(t) + jv_Q(t) va(t) ev(t) ≡ av(t)ejφv(t) ≡ Re(.) ×e−2πjf0t ×e+2πjf0t ⊗ δ(t) + _πtj vI(t) cos(.)− vQ(t) sin(.) 55 / 495

Theoretical determination of Rice components III

cos (2πf

0

t)

sin (2πf

0

t)

+

π₂

v

I

(t)

v (t)

v

Q

(t)

+

−

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Theoretical determination of Rice components IV

Any bandpass signal can be seen as the sum of two modulated signals by lowpass signals.

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A practical method to compute Rice components I

Theoretically,

+

π₂

v (t)

cos (2πf

₀

t)

Hilbert

sin (2πf

0

t)

+

v

I

(t)

(30)

A practical method to compute Rice components II

v (t)_{× 2 cos(2πf}0t) = 2 [vI(t) cos(2πf0t)− vQ(t) sin(2πf0t)] cos(2πf0t) = 2vI(t) cos2(2πf0t)− vQ(t) sin(2πf0t) cos(2πf0t)

= vI(t) + vI(t) cos(4πf0t)− vQ(t) sin(4πf0t) (67)

And after a low-pass filter,

v (t)_{× 2 cos(2πf}0t) −→ lowpass filter −→ vI(t) (68)

Likewise,

v (t)_{× 2 sin(2πf}0t) −→ lowpass filter −→ −vQ(t) (69)

59 / 495

A practical method to compute Rice components III

+

π₂

v

I

(t)

2 sin (2πf

0

t)

2 cos (2πf

0

t)

−v

Q

(t)

v (t)

(31)

Amplitude/phase representation of the complex envelope I

ev(t) = vI(t) + jvQ(t) = av(t) ejφv(t) (70) v(t) v_I(t) + jv_Q(t) va(t) ev(t) ≡ av(t)ejφv(t) ≡ Re(.) ×e−2πjf0t ×e+2πjf0t ⊗ δ(t) + _πtj vI(t) cos(.)− vQ(t) sin(.) 61 / 495

Amplitude/phase representation of the complex envelope II

By substituting ev(t) by its amplitude + phase description, we have

v (t) = Re ev(t)e2πjf0t (71) = Reav(t) ejφv(t)e2πjf0t (72) = av(t) cos (2πf0t + φv(t)) (73) with, by definition, av(t) = q v_I2(t) + v_Q2(t) (74) φv(t) = tan−1 vQ(t) vI(t) (75)

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Amplitude/phase representation of the complex envelope

III

Conclusions for v (t) = av(t) cos (2πf0t + φv(t))

1 the amplitude a_v(t) of the complex envelope is the envelope

of the original signal v (t).

2 the instantaneous phase of v (t) is given by the phase of the

complex envelope.

Any bandpass signal can be seen as a signal modulated both in amplitude and in phase.

63 / 495

Linear bandpass systems

Assume a bandpass filter, around f0, whose response filter is

h(t) = Ree_h(t)e2πjf0t ₍₇₆₎

The filtered signal is given by

y (t) = v (t)_{⊗ h(t)} (77) = Z ₊_∞ −∞ h(λ)v (t − λ)dλ (78) Thesis: ey(t) = 1 2 eh(t)⊗ ev(t) (79)

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Linear bandpass systems

Assume a bandpass filter, around f0, whose response filter is

h(t) = Reeh(t)e2πjf0t

(76) The filtered signal is given by

y (t) = v (t)_{⊗ h(t)} (77) = Z ₊_∞ −∞ h(λ)v (t − λ)dλ (78) Thesis: ey(t) = 1 2 eh(t)⊗ ev(t) (79) 65 / 495

Proof I

y (t) = v (t)_{⊗ h(t) =} Z +∞ −∞ h(λ)v (t _{− λ)dλ} (80) To get rid of Re (), we use

Re (a + jb) = a + jb 2 + a_{− jb} 2 = a + jb 2 + a + jb 2 ∗ (81) So, v (t _{− λ) =} 1₂ ev(t − λ)e2πjf0te−2πjf0λ + ev∗(t − λ)e−2πjf0te2πjf0λ

and h(λ) = 1₂ eh(λ)e2πjf0λ + e_h∗(λ)e−2πjf0λ Therefore, y (t) = 1 4e 2πjf0t Z +∞ −∞ eh(λ)ev(t − λ)dλ + 1 4e −2πjf0t Z +∞ −∞ e_h∗(λ)e_v∗(t _{− λ)}dλ + 1 4e −2πjf0t Z +∞ −∞ eh(λ)ev∗(t − λ)e4πjf0λdλ + 1 4e 2πjf0t Z +_∞ −∞ e_h∗(λ)ev(t − λ)e−4πjf0λdλ

(34)

Proof II

y (t) = 1 2Re Z +∞ −∞ eh(λ)ev(t − λ)dλe2πjf0t (82) + 1 2Re R+∞ −∞ eh∗(λ)ev(t − λ)e−4πjf0λdλe2πjf0t (83)

But, e_h∗(λ)ev(t − λ) is low frequency and e−4πjf0λ is high

frequency. Therefore, Z ₊_∞ −∞ e ∗ h(λ)ev(t − λ)e−4πjf0λdλ ≈ 0 (84) Finally, y (t) = 1 2Re ((eh(t) ⊗ ev(t)) e 2πjf0t₎ ₍₈₅₎ or ey(t) = 1 2 eh(t)⊗ ev(t) (86) 67 / 495

Bandpass filtering

Use of ey(t) = 1₂ eh(t)⊗ ev(t)

I the output is a bandpass signal too (_{⇒ it has a baseband} equivalent).

I we have a way to filter a bandpass signal by its baseband equivalent. This is really helpful when we work with digital signals (because the sampling frequency is much lower).

(35)

Analytic signal of a stochastic process

Two signals can be considered:

1 the stochastic process: X (t)

2 its auto-correlation function Γ_XX (τ ) or its power spectrum

γ_X(f )

The theory of equivalent baseband signals applies both to X (t) and to ΓXX (τ ), but they are very different concepts.

Remember that when a stochastic process X (t) passes through a (linear) filter, its power spectrum is multiplied by _{kH(f )k}2.

69 / 495

Link between a stochastic process and its complex envelope

X (t) is related to its complex envelope by X (t) = ReeX(t) e2πjf0t

(87) X (t) being a stochastic process, its complex envelope eX(t) is also

a stochastic process.

In general, X (t) is not stationary because its mean is time dependent.

Solution: introduction of a random phase Θ uniformly distributed over [0, 2π[

X (t) = ReeX(t) ej(2πf0t+Θ)

(36)

Rice decomposition of a stochastic process I

Likewise to developments for deterministic signals, the in-phase and quadrature components of a stochastic process can be expressed as eX(t) = XI(t) + j XQ(t) (89)

Rice components of the X (t) stochastic process are then obtained by X (t) = ReeX(t) e2πjf0t (90) = Re(XI(t) + j XQ(t)) e2πjf0t (91) = XI(t) cos(2πf0t)− XQ(t) sin(2πf0t) (92)

By analogy with the developments for deterministic signals, we build the analytic signal by filtering X (t) with the following filter H(f ) that removes all components for negative frequencies:

H(f ) =

(

0 if f < 0

2 if f _{≥ 0} (93)

71 / 495

Rice decomposition of a stochastic process II

By applying Wiener-Kintchine, the power spectral density of the analytic signal is given by

γ_X_a(f ) = _{kH(f )k}2γ_X(f ) (94) =

(

4 γX(f ) if f ≥ 0

0 if f < 0 (95) Note that it can also be written as γXa(f ) = 2H(f ) γX(f ).

(37)

Power spectrum

γ_e_X(f ) = γXa(f + f0) (96) ΓXX (τ ) = 1 2Re (ΓXaXa (τ )) (97) = 1 2Re ΓeXeX (τ ) e 2πjf0τ ₍₉₈₎

It is shown in a later chapter that, after stationarization, we get

γ_X(f ) = γeX(f − f0) + γ

∗

eX(−f − f0)

4 (99)

Important practical result: γX(f ) can be derived from γeX().

73 / 495

Outline

1 Reminder

5 Spread spectrum

interference

8 Multiplexing

(38)

Steps

1 Understand the origin of noise (thermal noise)

2 Look for a model of noise (in terms of the power spectral

density of a stochastic voltage)

3 Find a corresponding model for the noise getting out of a

one-port circuit (1₂kBT )

4 Define a practical way to derive the amount of noise when

dealing with a two-port circuit (noise figure, F0)

5 Find formulas to calculate the amount of noise accumulated

by the different elements of a cascade of two-port circuits (F0

for the cascade)

6 Derive good engineering rules for the order of elements in a

cascade

75 / 495

Noise in telecommunications systems I

Noise is a recurrent issue/problem in telecommunications systems. Remember for example the following theorem:

Theorem (Shannon-Hartley)

Channel capacity C (conditions for the error rate Pe → 0)

C _b s = B log₂ 1 + S N (100) where

I B is the channel bandwidth in Hz

I S

N the signal-to-noise ratio (in watts/watts, not in dB).

(39)

Noise in telecommunications systems II

There are several physical sources of noise:

I thermal noise

I shot noise

I ...

There are also “system” sources of noise

I quantization noise I intermodulation noise I crosstalk I interference noise I ... 77 / 495

Noise in telecommunications systems III

Approach to deal with noise in telecommunications systems?

1 model the most common noise: thermal noise for electronic

circuits

2 the principles of thermal noise can also be used to model

(40)

Modeling noise in electronic circuits

1 Characterization of a one-port circuit (source) Available noise power

Noise temperature of a one-port circuit Signal to noise ratio

2 Characterization of a two-port circuit (channel, amplifiers,

filters, etc)

Gain

Noise factor Figure of merit

Effective noise temperature Particular case: attenuator

3 Cascade of two-port circuits (a complete chain)

79 / 495

Thermal noise: a measuring experiment

R E (t)

(a) (b)

I (t) = E (t)_R noisy R

Figure: (a) Physical circuit with noise, (b) Th´evenin equivalent circuit for a resistor considered as a noise generator.

A conductive element with two terminals (_{≡ one-port circuit) may be} characterized by:

I its resistance R.

I Free electrons have some random motion depending on the temperature T _⇒ noise voltage source E (t).

(41)

Thermal noise: towards a model for E(t) I

A natural source of noise is thermal noise, caused by the omnipresent

motion of free electrons in conducting material.

I Because a voltage that would be measured at the output of a

resistance R is produced by many free electrons, by the central limit theorem, the amplitude statistics are modeled by a Gaussian.

The thermal noise has a zero mean (can be shown analytically and experimentally).

It can be shown that the autocorrelation function of thermal noise is well modeled by

ΓEE (τ ) = kBTRe −|τ|/t₀ t0 (101) where kB = 1.38× 10−23 [J/K] is Boltzmann’s constant.

T = 273.15 + C is the absolute temperature of the resistor (in Kelvin); C is the temperature in Celsius.

t0 = 10−12[s] is the average time between collisions of electrons.

81 / 495

Thermal noise: towards a model for E(t) II

As the autocorrelation function is

ΓEE (τ ) = kBTRe −|τ|/t0 t0 (102) the PSD is γ_E(f ) = 2kBTR 1 + (2πft0)2 (103)

Empirically, at room temperature, for low frequencies

(< 1000 [GHz]), the noise PSD is almost flat so that we may take

γ_E(f ) _{' 2k}BTR (104)

γ_E(f ) corresponds to the power of E (remember that the power is in the form of E2(t)).

(42)

Thermal noise: towards a model for E(t) III

Note: for a B bandwidth, a resistor in a short circuit dissipates a noise power of (non sinusoidal signal):

P = Z _∞ −∞γE(f )df /R = 2 Z _∞ 0 γ_E(f )df /R = 4kBT B (105)

Noise power spectral density

Consider a thermal noise with a noise temperature of T = 290 [K] and R = 1 [Ω], then we have

γ_E(f ) = 2 _{× 1.38 × 10}−23 _{× 290 × 1 = 8 × 10}−21 W Hz (106) Note that γE(f ) is the power due to thermal noise; it is not the

power available at the output!

In the following, we calculate the available noise power.

83 / 495

Characterization of a single-port circuit (dipole)

Source impedance: Zs (f ) = Rs (f ) + jXs (f ) (107) Load impedance: ZL(f ) (108)

Z

s

Z

L

E

(43)

Available power

Definition (Available power)

The available power is the maximum power that can be drawn from a source.

Theorem (Maximum power transfer)

The maximum power transfer occurs when the load impedance is equal to the conjugate of the source impedance (matched

impedances):

ZL(f ) = Z_S∗(f ) (109)

If impedances are almost purely resistive, then

RL = RS (110)

85 / 495

The case of sinusoidal signals I

When the signals are sinusoidal, the power provided by a source S at the output of the dipole, PpS, is given by

P_pS = lim T_→∞ 1 T Z _T 0 v (t)i(t)dt (111) = 1 2Re b VbI∗ (112)

where V andb bI are phasors are defined with peak values, instead of root mean square (rms) values, and therefore there is a 1₂ factor.

(44)

The case of sinusoidal signals II

Zs ZL E I V

In a load ZL, we have (voltage divider): b V = ZL Zs + ZL b E and bI = Eb Zs + ZL Therefore, PpS = 1 2Re b VbI∗ = 1 2Re ZLEb Zs + ZL b E∗ (Zs + ZL)∗ ! = Re (ZL) bE 2 2_kZs + ZLk2 (113) 87 / 495

The case of sinusoidal signals III

So, the available power (that is when ZL(f ) = Z_S∗(f ), so that

Zs + ZL = 2 Re (Zs)) from the Source is PaS = Re (ZL) bE2 2_kZs + ZLk2 = Re (ZL) bE 2 8 Re (Zs)2 = Eb 2 8 Re (Zs) (114)

(45)

The case of sinusoidal signals IV

Summary

The available power in the load is PaS =

b E2 8 Re (Zs)

(115) By definition, the effective power produced by the source is (for an open circuit, matched impedance, so that Zs + ZL = 2 × Re (Zs))

PS = 1 2EbIb ∗ ₌ 1 2Eb b E∗ (Zs + ZL)∗ = Eb 2 4 Re (Zs) (116)

It follows then that the power delivered by the Th´evenin generator and the power dissipated in the generator’s Th´evenin resistance are the same.

89 / 495

Available noise power (matched impedance)

For an arbitrary random noise source, the provided noise power is PpN = lim T_→+∞ 1 T Z _T 0 V (t)I (t)dt (117) R E R _I V

For the particular case of thermal noise

The available power spectral density is, by applying Wiener-Kintchine: γ_aN(f ) = _{kH(f )k}2γ_E(f ) = R R + R 2 2kBT = kBT 2 (118)

Remember that this assumes the adaptation of the impedance (Zs = Z_L∗); otherwise, the result would depend on the impedances!

(46)

Noise at the output of an antenna pointing towards the sky

Sky temperature R γE(f ) = 2kBTa Ta Noiseless resistor Noise source Example Contributions to Ta:

I the sky contributes to 10 [K] (3 [K] of residual temperature of the Big Bang + 7 [K] due to atmospheric absorption).

I the ground contribution is typically 0.1 of 290 [K] (contributions of secondary lobes looking at the ground)

Thus,

Ta = 0.9× 10 +0.1× 290 = 38 [K] (119)

The noise PSD is then

γaN(f ) = kBT 2 = 2.6 × 10 −22 hW Hz i (120) 91 / 495

A note on matched impedances

There exist two ways to match loads:

1 Z∗

S = ZL (conjugate matching). This ensures the maximum

transfer of power (that is the “available power”).

2 Z_c = Z_L, where Z_c denotes the characteristic impedance of a

transmission line. Zc is the ratio of the amplitudes of voltage

and current of a single wave propagating along the line. When Zc = ZL, there is no reflection.

In practice, what should we do when we connect a circuit to a line?

I Z_S∗ = ZL is mandatory. It is the matching condition for this

chapter.

I Luckily, Zc = Rc + jLc and Lc Rc, so that Zc ' Rc.

(47)

Summary I

Arbitrary load Matched load

Sinusoidal signals PpS = 1₂Re b VbI∗ = bE2Re(ZL) 2kZs+ZLk2 PaS = bE2 8Re(Zs)

Stochastic processes PpN = limT→+∞ _T1 R₀T V (t)I (t)dt γaN(f ) = γE(f ) 4Re(Zs)

Thermal noise γaN(f ) = kB₂T

Table: Power provided by a one-port circuit.

Theorem (Available power)

The available power from a thermal source in a bandwidth of B is

PN = Z ₊_∞ −∞ γaN(f ) df = 2× Z _f₀₊B 2 f0−B₂ kBT 2 df = kBTB (121) Conclusions: 93 / 495

Summary II

We have a model for the power of (thermal) noise

P_N = k_BTB (122)

It does not depend on the impedance (because we assume impedance matching). It depends only on

I the temperature T

I the bandwidth B

Noise power

Consider a thermal noise with a noise temperature of T = 290 [K] and B = 10 [MHz], then we have

PN(f ) = 1.38× 10−23 × 290 × 107 = 40 × 10−15 [W] (123)

(48)

Summary III

PN = kBTB (124)

So the noise power increases with the bandwidth _{⇒ we may want} to reduce the bandwidth

However, the capacity is given by C = B log₂ 1 + S N (125) If S is fixed and we only have a white noise, then

C = B log₂ 1 + S kBTB (126) To increase the capacity C , there is no alternative other than to increase the bandwidth _{⇒ trade-off !}

95 / 495

Noise temperature for a one-port circuit [generic model]

Definition (Noise temperature [at a given frequency])

The noise temperature at a given frequency is the absolute temperature that an impedance should have to produce, by

thermal effect, for a given frequency, a noise power spectral density equal to that of the circuit.

By definition thus,

γ_aN(f ) = kBT (f )

2 (127)

Definition (Frequency average for T (f ) and bandwidth)

The maximum temperature, denoted T, the noise temperature of the dipole and the bandwidth are defined such that

(49)

Noise temperature for a one-port circuit [generic model]

Definition (Noise temperature [at a given frequency])

The noise temperature at a given frequency is the absolute temperature that an impedance should have to produce, by

thermal effect, for a given frequency, a noise power spectral density equal to that of the circuit.

By definition thus,

γ_aN(f ) = kBT (f )

2 (127)

Definition (Frequency average for T (f ) and bandwidth)

The maximum temperature, denoted T, the noise temperature of the dipole and the bandwidth are defined such that

PaN = kBTB (128)

97 / 495

Signal to noise ratio of a one-port circuit

Definition

The Signal to Noise ratio (S/N) of a dipole is defined as the ratio between the available power of the signal and the noise power

S

N =

PaS

P_aN (129)

Warning: in the following S denotes the Signal and not the noise source!

By convention, if the signal is modulated, the power of the useful signal is defined as follows (it is just a reference to compare

techniques):

I for amplitude or angular modulations, we consider the power of the carrier (it is then a carrier to noise ratio _NC),

I for suppressed carrier amplitude modulation techniques, we take the mean power of the modulating signal, and

(50)

Characterization of a two-port circuit (quadripole) I

1

1 ₂

2 linear

quadripole

output

Z

S

E

V

input

Figure: Scheme of a two-port circuit.

99 / 495

Characterization of a two-port circuit (quadripole) II

Steps:

I Notion of gain?

I Characterization of the amount of internal noise of the

quadripole by means of the normalized notion of noise figure

I Equivalent circuits

I Figure of merit (not normalized)

I Effective noise temperature

I Special case: purely resistive attenuator

Hypothesis: loads are matched (complex conjugate) at the input and at the output

(51)

Noise figure of a two-port circuit

Definition (Noise Figure (NF), F0)

Assuming fixed internal impedances, the spot noise figure of a two-port circuit, for a given frequency f , denoted F0(f ), is the

ratio between

(1) the noise power spectral density at the output of the quadripole, for the appropriate frequency f , when the noise temperature of the generating one-port put at the input is normalized to be T0 = 290 [K ], and

(2) the contribution of the generating input source, at a frequency f , to the noise power spectral density at the output.

1 1 ₂ 2 quadripole Rs Rin= Rs γaN1(f ) = 1 2kBT0 γaN2(f ) = G (f )γaN1(f ) + γaNq(f ) linear 4kBTRsB 101 / 495

Property: F

0

>

1

1 1 ₂ 2 quadripole Rs Rin = Rs γaN1(f ) = 1 2kBT0 γaN2(f ) = G (f )γaN1(f ) + γaNq(f ) linear 4kBTRsB So, by definition, F0(f ) = γ_aN₂(f ) G (f )γaN1(f ) = G (f )γaN1(f ) + γaNq(f ) G (f )γaN1(f ) > 1 (130)

(52)

Equivalent circuit

F0(f ) = G (f )γaN1(f ) +γaNq(f ) G (f )γaN1(f ) ⇐⇒ γaNq(f ) = [(F0(f )− 1) γaN1(f )] G (f ) G G γaN2(f ) = G (f )γaN1(f ) + γaNq(f ) γaN1(f ) γaN1(f ) γaN2(f ) = G (f )F0(f )γaN1(f ) (F0(f )− 1)γaN1(f ) γaNq(f )

Figure: Scheme of a noisy quadripole and an equivalent circuit (we model the internal noise by (F0(f )− 1) γaN1(f ) at the entrance)

103 / 495

Interpretation of the notion of noise figure

The input signal to noise ratio is given by

S N in = γin(f ) γ_aN₁(f ) (131) At the output of the two-port circuit, we have

_S N out = γout(f ) γ_aN₂(f ) (132) Therefore, S N in S N out = γin(f ) γ_aN₁(f ) γ_aN₂(f ) γout(f ) (133)

As _{γout(f ) = G(f )γin(f )} , this ratio becomes

S N in S N out = γaN2(f ) γ_aN₁(f )G (f ) = G (f )F0(f )γaN1(f ) γ_aN₁(f )G (f ) = F0(f ) (134)

(53)

Noise figures for un-normalized temperatures (

T

s

6= T

0

)

What if the input temperature Ts 6= T0?

This leads to define a different notion _{⇒ the} figure of merit F. Link between F and F0 (F0 is provided by the manufacturer)?

Remember that, considering γaN1(f )|T =T0 =

1

2kBT0:

γ_aNq(f ) = (F0 − 1) 1

2kBT0G (f ) (135) The internal noise of a two-port circuit is independent of the input temperature (the last is just a convention). Therefore, another temperature Ts then leads to another figure of merit F . It is

derived as follows γ_aNq(f ) = (F0 − 1) 1 2kBT0G (f ) = (F − 1) 1 2kBTsG (f ) (136) and, finally, F = 1 + T0 Ts (F0 − 1) (137) 105 / 495

Effective noise temperature of a two-port circuit I

G G γaN2(f ) = G (f )γaN1(f ) + γaNq(f ) γaN1(f ) γaN1(f ) γaN2(f ) = G (f )F0(f )γaN1(f ) (F0(f )− 1)γaN1(f ) γaNq(f ) γ_aN₂(f ) = 1 2kBT0G (f ) + γaNq(f ) = 1 2kB [T0 +(F0 − 1)T0] G (f )

(54)

Effective noise temperature of a two-port circuit II

Definition (Effective (input-)noise temperature)

Te = (F0 − 1)T0 (138)

It is the additional temperature required for an input source to produce the same available power at the output.

Note that:

Te = (F0 − 1)T0 ⇔ F0 = 1 + Te

T0

(139)

Noisy two-port circuit

Consider an effective noise temperature Te = 120 [K], then

F0 = 1 + 120

290 = 1.41 = 1.5 [dB] (140)

107 / 495

Attenuator with a “Gain” G = 1/L or Loss L

Let us take an attenuator (for example a purely resistive circuit or a lossy transmission line) at temperature T0.

(1) Assuming matched input and output impedances, the available noise power at the output is

γ_aN₂(f ) = 1

2kBT0 (141)

(2) But that the attenuator is characterized by its effective noise temperature Te, then the output noise power is

γ_aN₂(f ) = 1

2kB (T0 + Te) 1

L (142)

By combining these two expressions: Te = (L − 1)T0. So that,

F0 = 1 + (L− 1)T0 T0

= L (143)

Conclusion: attenuators have a noise figure F0 equal to their

(55)

Attenuator with a “Gain” G = 1/L or Loss L

Let us take an attenuator (for example a purely resistive circuit or a lossy transmission line) at temperature T0.

(1) Assuming matched input and output impedances, the available noise power at the output is

γ_aN₂(f ) = 1

2kBT0 (141)

(2) But that the attenuator is characterized by its effective noise temperature Te, then the output noise power is

γ_aN₂(f ) = 1

2kB (T0 + Te) 1

L (142)

By combining these two expressions: Te = (L − 1)T0. So that,

F0 = 1 +

(L_{− 1)T}0

T0 = L (143)

Conclusion: attenuators have a noise figure F0 equal to their

attenuation ratio L when their physical temperature equals T0.

109 / 495

What about the attenuator L at other temperatures?

Let us consider Ts 6= T0 and calculate F .

There are two possible ways to calculate F :

1 same reasoning as previously: it is impossible to discriminate

the output of the two-port circuit from the input one-port circuit, so that γ_aN₂(f ) = 1 2kBTS = γaN2(f ) = 1 2kB (TS + Te) 1 L (144) and F = L.

2 Remember that F₀ represents the signal to noise degradation

F0 = S N in/ S N out = L (145) But, as L =Sin/Sout, we have Nin = Nout.

Conclusion: for an attenuator with a factor L, the amount of noise is always unaffected, so that

(56)

Noise figure of two-port networks I

γaN₁(f ) (F01− 1)γaN1(f ) (F02− 1)γaN₁(f ) G1F01γaN₁(f ) G2 G1 F02 F01

Figure: Cascading two-port elements.

For a two-port network with 2 stages, F0 = γ_aN₁(f )G1G2 + (F01 − 1) γaN1(f )G1G2 + (F02 − 1) γaN1(f )G2 γ_aN₁(f )G1G2 = 1 + (F01 − 1) + (F02 − 1) G1 (147) = F01 + (F02 − 1) G1 (148) 111 / 495

Noise figure of two-port networks II

For a two-port network with n stages, F0 = F01 + F02 − 1 G1 + F03 − 1 G1G2 +· · · = F01 + n X i=2 F0i − 1 Q_i₋₁ j=1Gj (149) Likewise, Te = Te1 + Te2 G1 + Te3 G1G2 +· · · = Te1 + n X i=2 Tei Q_i₋₁ j=1Gj (150)

(57)

Low-noise amplifier for receivers

F0 = F01 + F02 − 1 G1 + F03 − 1 G1G2 + · · · (151) Two consequences:

1 the noise figure always increases with an additional stage. 2 the overall noise figure of a receiver is primarily set by the

noise figure of its first amplifying stage.

Therefore, the first stage amplifier is often a Low-Noise Amplifier (LNA). Then, the overall receiver noise figure is

Freceiver ' FLNA + Fothers − 1_GLNA (152)

113 / 495

Outline

1 Reminder

5 Spread spectrum

interference

8 Multiplexing

(58)

Digital modulation

1 Criteria ruling the selection of a modulation scheme 2 Definition and typology of digital modulations

3 Classic linear modulations Description

Determination of the power spectrum Amplitude modulation (ASK)

Phase modulation (PSK)

Quadrature modulation (QPSK) 4 Offset modulations

Description

Determination of the power spectrum Offset quadrature modulation (OQPSK)

Minimum shift modulation techniques (MSK)

115 / 495

(59)

Criteria for choosing of a modulation scheme

1 Resistance to distortion and perturbation. That includes: resistance to additive white Gaussian noise. Noise usually results in a bit error rate Pe, expressed in terms of the Eb/N0

ratio.

sensitivity to interference (multipath, other users, etc). sensitivity to imperfect filters. This is associated to the phenomenon of intersymbol interference.

sensitivity to non-linearities. 2 Spectral occupancy:

1 spectral efficiency η, expressed in bit per second per Hertz

h

b/s Hz

i

, which measures the bit rate that can be transmitted per unit of frequency bandwidth for a given modulation.

2 asymptotic behavior, defined by the values of the spectral density for frequencies relatively distant from the carrier frequency.

3 Simplicity of implementation.

117 / 495

Spectral efficiency of linear modulation techniques

C = B log₂1 + EbRb B N0 _{max: R} b→C =_⇒ η _' Rb B = log2 1 +ηEb N0

(60)

Understanding digital modulations

We will make an intensive use of alternative representations

v(t) vI(t) + jvQ(t) va(t) ev(t) ≡ av(t)ejφv(t) ≡ Re(.) ×e−2πjf0t ×e+2πjf0t ⊗ δ(t) + _πtj vI(t) cos(.)− vQ(t) sin(.) 119 / 495

Reminder: building a digital signal

P+∞_k=_−∞

A

k

p(t

− kT )

−1 −1

+1

Information: Pulse shape:

pulse shapes Combining modulated Result: A0 A1 A2 Ak p(t) t t t p(t− kT ) p(t_{− T )} p(t_{− 2T )} −T /2 T /2 T t t t t A1p(t− T ) A0p(t) A2p(t− 2T ) +_X∞ k=−∞ Akp(t− kT )

Main characteristics of digital signals:

1 With digital signals, the fundamental unit is a time slot T . 2 One information symbol A_k per time slot T (no overlap).

p(t _{− kT ) thus also acts as a} time window!

3 Same pulse shape for each time slot (eases the task of the

(61)

Reminder: building a digital signal

P+∞_k=_−∞

A

k

p(t

− kT )

−1 −1

+1

Information: Pulse shape:

pulse shapes Combining modulated Result: A0 A1 A2 Ak p(t) t t t p(t_{− kT )} p(t_{− T )} p(t_{− 2T )} −T /2 T /2 T t t t t A1p(t− T ) A0p(t) A2p(t− 2T ) +_X∞ k=−∞ Akp(t− kT )

Main characteristics of digital signals:

1 With digital signals, the fundamental unit is a time slot T . 2 One information symbol A_k per time slot T (no overlap).

p(t _{− kT ) thus also acts as a} time window!

3 Same pulse shape for each time slot (eases the task of the

receiver).

121 / 495

Definition and typology of digital modulations I

Definition (General expression for digital modulations (based on the complex envelope))

s(t) = Reψ [m(t)]ej(2πfct+ϕc) ₍₁₅₃₎

The complex function ψ [m(t)], which is related to the modulating waveform m(t), defines the type of modulation. It is also the

complex envelope es(t) of the modulated signal s(t).

Depending on the form of ψ(.) = ψI(.) + j ψQ(.), we generally

distinguish:

I linear modulations for which ψ [m(t)] is a linear function of m(t).

I angular modulations for which ψ [m(t)] has the form of

ψ [m(t)] = ejϕ[m(t)] (154) where ϕ [m(t)] is a linear function of m(t).

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Definition and typology of digital modulations II

The modulated signal can also be expressed as

s(t) = ψ_I [m(t)] cos (2πfct + ϕc)− ψQ [m(t)] sin (2πfct + ϕc)

(155) and

s(t) = _{kψ [m(t)]k cos (2πf}ct + ϕc + arg ψ [m(t)]) (156)

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Definition and typology of digital modulations III

In the following, we will concentrate on modulations that can be written as s(t) = Re     +_X_∞ k=_−∞ dk(t) ej(θk−2πfckT )  _ej(2πfct+ϕc)   ₍₁₅₇₎

Two types of linear modulations will be studied:

1 “classic”modulations, for which θ_k = 2πf_ckT 2 offset modulations, for which θ_k = 2πf_ckT + kπ

2

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Classic linear modulations

Description

Classic linear modulations are such that θ_k = 2πfckT. Therefore,

we have that

s(t) = Rees(t) ej(2πfct+ϕc)

(158) The complex envelope takes the form

es(t) = +∞ X k=_−∞ dk(t) (159) = +∞ X k=_−∞ Dk pk(t − kT ) (160)

where Dk=Ak+jBk is complex and Ak, Bk are two real random

variables.

In most cases, the pulse shape of pk(t − kT ) is the same for each

symbol k. Therefore, pk(.) becomes p().

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Complex envelope of classic linear modulation techniques

The complex envelope can also be written as es(t) = sI(t) + jsQ(t), so that sI(t) = +∞ X k=−∞ Ak p(t − kT ) (161) sQ(t) = +∞ X k=_−∞ Bk p(t − kT ) (162) resulting in s(t) = sI(t) cos (2πfct + ϕc)− sQ(t) sin (2πfct + ϕc) (163)

and, by replacing sI and sQ by their value,

s(t) = " ₊_∞ X k=−∞ Ak p(t− kT ) # cos (2πfct + ϕc)− " ₊_∞ X k=−∞ Bk p(t− kT ) # sin (2πfct + ϕc)

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Derivation of the power spectral density I

Power spectral density of a modulated signal?

The modulated signal is a stochastic process S(t) that can be written, taking ϕc = 0, as

S(t) = ReM(t) ej2πfct ₍₁₆₄₎

where M(t) is a complex stochastic process (such as the complex envelope es(t) in our case).

But is S(t) stationary?

Obviously, we take M(t) stationary. However, even then

µ_S = E _{{S(t)} = µ}M ej2πfct (165)

is time-dependent (not constant).

Therefore, as such, we cannot calculate γS(f ).

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Derivation of the power spectral density II

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Derivation of the power spectral density III

Stationarization

The S(t) process is not stationary because its mean is time-dependent. We have to stationarize the signal.

For that purpose, we add the random phase Θ whose probability density function (pdf) is uniformly distributed over [0, 2π[ (in other words, pdf_Θ(θ) = _2π1 for θ _{∈ [0, 2π[ and 0 outside):}

S(t) = ReM(t) ej(2πfct+Θ) ₍₁₆₆₎

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Derivation of the power spectral density IV

Mean of S(t) = ReM(t) ej(2πfct+Θ)?

Note that M(t) and Θ are independent. For the computation, we do

1 Re (a + jb) is replaced by Re (a + jb) = a+jb

2 +

a_−jb 2

2 Then, we take the expectation of both terms: the expectation

of a sum is the sum of the expectations.

3 Then, for the first term, we have

1 2E n M(t) ej(2πfct+Θ)o ₌ 1 2E {M(t)} E n ej(2πfct+Θ)o = 1 2µM Z _2π 0 e j(2πfct+Θ) 1 2πdθ = 1 2µM × 0 = 0 µ_S = E _{{S(t)} = 0} (167)

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Derivation of the power spectral density V

Autocorrelation function of S(t)? ΓSS (t, t − τ) = E {S(t) S(t − τ)} (168) As S(t) = ReM(t) ej(2πfct+Θ) ₍₁₆₉₎ = 1 2 h M(t) ej(2πfct+Θ) _{+ M}∗_{(t) e}−j(2πfct+Θ)i ₍₁₇₀₎ we have S(t)S(t _{− τ}) = 1 2 h M(t) ej(2πfct+Θ) _{+ M}∗_{(t) e}−j(2πfct+Θ)i ₍₁₇₁₎ × 1 2 h M(t _{− τ) e}j(2πfc(t−τ)+Θ) _{+ M}∗_(t _{− τ) e}−j(2πfc(t−τ)+Θ)i 131 / 495

Derivation of the power spectral density VI

S(t)S(t _{− τ}) = 1 4M(t) e j(2πfct+Θ)_M(t _{− τ) e}j(2πfc(t−τ)+Θ) ₍₁₇₂₎ + 1 4M(t) e j(2πfct+Θ)_M∗_(t _{− τ) e}−j(2πfc(t−τ)+Θ)₍₁₇₃₎ + 1 4M ∗_{(t) e}−j(2πfct+Θ)_M(t _{− τ) e}j(2πfc(t−τ)+Θ)₍₁₇₄₎ + 1 4M ∗_{(t) e}−j(2πfct+Θ)_M∗_(t _{− τ) e}−j(2πfc(t−τ)+Θ) S(t)S(t _{− τ}) = 1 4M(t)M(t − τ)e j(2πfc(2t−τ)+2Θ) ₍₁₇₅₎ + 1 4M(t)M ∗_(t _{− τ)}_ej2πfcτ ₍₁₇₆₎ + 1 4M ∗_(t)_M(t _{− τ)}_e−j2πfcτ ₍₁₇₇₎ + 1 4M ∗_(t)_M∗_(t _{− τ)}_e−j(2πfc(2t−τ)+2Θ) ₍₁₇₈₎

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Derivation of the power spectral density VII

Because, ΓSS (t, t − τ) = E {S(t)S(t − τ)}, the terms of E {.}

containing 2Θ are null. Therefore, we are left with

ΓSS (t, t − τ) = 1 4E M(t) M∗(t _{− τ) e}j2πfcτ _{+ M}∗_{(t) M(t} − τ) e−j2πfcτ = 1 4E 2 Re M(t) M∗(t _{− τ) e}j2πfcτ ₍₁₇₉₎ = 1 2Re E M(t) M∗(t _{− τ) e}j2πfcτ ₍₁₈₀₎ = 1 2Re ΓMM (t, t − τ) e j2πf_cτ ₍₁₈₁₎ Finally, because ΓSS (τ ) = 1 4 ΓMM (τ ) ej2πfcτ + ΓMM(τ )∗ e−j2πfcτ (182)

and x∗(t) _{↔ X}∗(_{−f ), we have that}

γS(f ) = γM (f − fc) + γ ∗

M (−f − fc)

4 (183)

where γM(f ) is the power spectral density of M(t).

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Derivation of the power spectral density VIII

Power spectral density of a modulation signal for classic linear modulation techniques If M(t) is real: γ_S(f ) = γM (f − fc) + γM (f + fc) 4 (184) If M(t) is complex: γS(f ) = γ_M (f _{− f}c) + γ_M∗ (−f − fc) 4 (185)

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Calculation of the power spectrum of the complex envelope

(for classic linear modulation techniques)

The complex envelope of the modulated signal is (es(t) = M(t))

M(t) = +∞

X

k=−∞

Dk p(t − kT ) (186)

The sequence of complex random variables Dk is characterized by

I mean: µD = E {Dk}

I variance: σD2 = E {(Dk − µD) (Dk − µD)∗}

I autocorrelation function: ΓDD (k, k − l) = E {DkDk∗−l}

I covariance function: CDD(k, k − l) = E {(Dk − µD) (Dk−l − µD)∗}

After the stationarization of Dk, the PSD of the baseband complex

envelope is (cf. first course in telecommunications)

γM(f )= kP(f )k 2 T " σ2D +kµDk2 +∞ X m=−∞ 1 Tδ f ₋ m T # (187)

In conclusion (γM(f ) being real in this case):

γS(f ) = γM (f − fc) + γM(f + fc)

4 (188)

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Amplitude Shift Keying (amplitude modulation) I

Definition (Complex envelope of the Amplitude Shift Keying (ASK))

es(t) =

+_X_∞ k=_−∞

Ak p(t − kT ) (189)

Common choice for the shaping pulse function over [0, T ]:

p(t) = rect_{[0,T ]}(t) (190)

Goal of the following slides: we need to find the envelope a(t) and the phase ϕ(t) of the modulated signal. These two signals can

be derived from

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Amplitude Shift Keying (amplitude modulation) II

Trick: _±Ak can be expressed as ±Ak = kAkk e(1−sign(Ak))

π

2j

Let us verify this:

I Ak is positive (Ak = kAkk): Ak = kAkk e(1−sign(Ak)) π 2j =kA_kk e(1−1)π2j = kA_kk e0π2j = kA_kk (192) I Ak is negative (Ak =−kAkk): Ak =kAkk e(1−(−1)) π 2j = kA_kk e2π2j =kA_kk eπj =−kA_kk (193) So, we have es(t) = +_∞ X k=−∞ Ak p(t − kT ) (194) = +∞ X k=−∞ kAkk e(1−sign(Ak)) π 2j p(t − kT ) (195) 137 / 495

Amplitude Shift Keying (amplitude modulation) III

Therefore, the envelope a(t) and the phase ϕ(t) of the modulated

signal are given by a(t) = +_X_∞ k=_−∞ kAkk rect[0,T ](t − kT ) (196) ϕ(t) = +_X_∞ k=_−∞ π 2 (1− sign (Ak)) rect[0,T ](t − kT ) (197)

Note the presence of the time windowing function

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Power spectral density of the ASK-2

Hypothesis: both signals _{±A have an equal probability!}

I The mean µA of the random variable Ak is equal to 0.

I The variance is given by σ_A2 = E A2_k = A2.

I p(t) = rect_{[0,T ]}(t). Thus, its Fourier transform is

P(f ) = e−j2πf T2 T sinc(fT ) (198)

Therefore, the power spectrum of the complex envelop is

γ_e_s(f ) = kP(f )k_T 2 hσ_A2 +_kµAk2P+_m=∞_−∞ _T1δ f − m_T i : γes(f ) = A 2_{T sinc}2_{(fT )} ₍₁₉₉₎

and that of the ASK-2 modulated signal is

γ_s(f ) = A

2_T _sinc2_[(f _{− f}

c) T ] + sinc2[(f + fc) T ]

4 (200)

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ASK-2: bandwidth and spectral efficiency

The PSD of ASK-2 is γ_s(f ) = A 2_T _sinc2_[(f _{− f} c) T ] + sinc2[(f + fc) T ] 4 (201) At fixed bitrate Rb:

Technique bandwidth spectral efficiency η Baseband (NRZ) W = 0.6 Rb η = _{0.6 R}Rb _b ' 1.6

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Constellation or state diagram: definition I

We have:

s(t) = sI(t) cos (2πfct)−sQ(t) sin (2πfct) (202) and, for the complex envelope:

es(t) = sI(t)+ jsQ(t) (203) The alternative is s(t) = as(t)cos (2πfct +φs(t)) (204) with as(t) = q s2 I (t) + sQ2(t) (205) and φs(t) = tan−1 sQ(t) sI(t) (206)

Definition (Constellation diagram)

The plot of es(t) in a diagram whose axis units are (cos (2πfct),

− sin (2πfct)) defines the constellation diagram.

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Constellation or state diagram: definition II

Constellation diagram of a ASK-2 or BPSK modulation

(A, 0) (_{−A, 0)}

−p(t) sin (2πfct + ϕc)

p(t) cos (2πfct + ϕc)

The presence of p(t) reinforces the idea that there is only one symbol per time slot, but it is often dropped. Likewise, ϕc is often taken equal to 0.

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Constellation diagram: purposes

(A, 0) (−A, 0)

p(t) cos (2πfct + ϕc)

I Representation of the possible states in the complex plane.

I See how the state diagram is used (here, we immediately see that the sin() axis is not used).

I The distance between states is essential for finding the Pe.

Closer states mean less resistance to noise.

I See the paths from one state to another (trajectories).

Note that we move from one state to another state at the rhythm of the symbol rate (not the bit rate!)

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Constellation diagram of an On-Off Shift Keying (OOK)

(A, 0)

p(t) cos (2πfct + ϕc) (0, 0)

Infrared signals are usually sent using On-Off Shift Keying (because it is hard to determine the phase of an infrared signal).

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Constellation diagrams

OOK, ASK-2_{≡BPSK≡PSK-2, and QPSK.}

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